This is the first problem from me

Calculus Level 2

Let f ( x ) = ∫ x 3 sin ⁡ t t d t \displaystyle f(x) = \int_x^3 \dfrac{\sin t}t \, dt , compute lim ⁡ x → 3 x 2 f ( x ) 3 − x \displaystyle \lim_{x\to3} \dfrac{x^2 f(x)}{3 - x} .

sin ⁡ 3 3 \dfrac{\sin3}3 sin ⁡ 3 \sin3 3 sin ⁡ 3 3\sin3 − sin ⁡ 3 3 -\dfrac{\sin3}3

ابو الحسن قاهر الاحباب by ابو الحسن قاهر الاحباب , 25, Jordan

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1 solution

I am sorry, I do not know what the correct answer, but I want to know what is the correct answer and why, please help me..

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Hint: Since the form is 0/0 you can apply L'hopital rule (For differentiating numerator use product rule.)

Rishabh Jain - 5 years, 5 months ago

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Note that f ( 3 ) = 0 f(3)=0 and f ′ ( 3 ) = − sin ⁡ 3 3 f'(3)=-\frac{\sin3}{3} by the fundamental theorem (reverse the limits of integration!). The limit we seek is ( lim ⁡ x → 3 x 2 ) ( lim ⁡ x → 3 f ( x ) − f ( 3 ) x − 3 ) = 9 f ′ ( 3 ) = − 3 sin ⁡ ( 3 ) (\lim_{x\to 3}x^2)\left(\lim_{x\to 3}\frac{f(x)-f(3)}{x-3}\right)=9f'(3)=\boxed{-3\sin(3)} (the given answer is missing a negative sign)

Otto Bretscher - 5 years, 5 months ago

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Thank you so much

ابو الحسن قاهر الاحباب - 5 years, 4 months ago

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